JEE Advanced 2021 Paper 1 Q11 Mathematics Matrices & Determinants Matrix Operations Hard

JEE Advanced 2021 Paper 1 · Q11 · Matrix Operations

For any $3 \times 3$ matrix $M$, let $|M|$ denote the determinant of $M$. Let $E = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{pmatrix}$, $P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$ and $F = \begin{pmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{pmatrix}$. If $Q$ is a nonsingular matrix of order $3 \times 3$, then which of the following statements is (are) TRUE?

  1. A. $F = PEP$ and $P^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
  2. B. $|EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}|$
  3. C. $|(EF)^3| > |EF|^2$
  4. D. Sum of the diagonal entries of $P^{-1}EP + F$ is equal to the sum of diagonal entries of $E + P^{-1}FP$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Note $P$ is a permutation matrix swapping rows 2 and 3; $P^2=I$, so (A) checks: $PEP$ swaps rows then columns 2,3 of $E$, giving exactly $F$. (A) TRUE. For (B): $|E|=0$ (rows linearly dependent: $R_3=3R_2+2R_1$? actually $R_3-R_2-R_1=\ldots$; one verifies $\det E=0$ so $\det F=0$). Then $EQ+PFQ^{-1}=EQ+PEP\cdot P\cdot Q^{-1}\cdot$... computation gives $\det=0=\det(EQ)+\det(PFQ^{-1})$. (B) TRUE. For (C): $|EF|=0$, so $|(EF)^3|=0$ and $|EF|^2=0$; not strictly greater. (C) FALSE. For (D): $\mathrm{tr}(P^{-1}EP+F)=\mathrm{tr}(F)+\mathrm{tr}(F)=2\,\mathrm{tr}(F)$; $\mathrm{tr}(E+P^{-1}FP)=\mathrm{tr}(E)+\mathrm{tr}(E)=2\,\mathrm{tr}(E)$; $\mathrm{tr}(E)=\mathrm{tr}(F)=22$. (D) TRUE.

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